Mathematics Test 149

Result

Mathematics Test 149

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Weekly Quiz Competition

Question 1
4 / -1

The set of all values of 'a' for which the equation cos 2x + a sin x = 2a – 7 has a solution is

A
(– ∞, 2)

B
[2, 6]

C
(6, ∞)

D
(– ∞, ∞)

Solution
Question 2
4 / -1

Negation of "If India wins the match then India will reach in the final" is :-

A
If India doesn't win match then India will not reach in the final.

B
India wins the match and India will not reach in the final.

C
India doesn't win the match and India will reach in the final.

D
None of these

Solution
Question 3
4 / -1

If ax² – bx + 5 = 0 does not have two different real roots, then the minimum value of 5a + b will be

A
-1

B
2

C
3

D
4

Solution

f(0) = 5 > 0 and D ≤ 0

so f(x) ≥ 0

f(–5) ≥ 0 ⇒ 5a + b ≥ –1

least value of 5a + b is –1
Question 4
4 / -1

Directions For Questions

Observe the numerical triangle as shown below.

...view full instructions

The first term in 15^throw will be

A
2^–100

B
2^–105

C
2^–110

D
2^–115

Solution
Question 5
4 / -1

|z – 2 – 3i|² + |z – 4 – 3i|² = λ represents the equation of a circle with least radius, then λ = ?

A
4

B
3

C
2

D
1

Solution
Question 6
4 / -1

Bag A contains 4 green and 3 red balls and Bag B contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted that it is green then the probability that is comes from Bag B is :-

A
2/3

B
2/7

C
3/7

D
1/3

Solution
Question 7
4 / -1

Directions For Questions

...view full instructions

and det(A) = det(4I), where I is 3 × 3 identity matrix, then (a – b)³ + (b – c)³ + (c – a)³ can be equal to -

A
-10

B
-24

C
-18

D
-6

Solution

⇒   det(A) = (a – b)² (b – c)² (c – a)²

      & det (4I) = 64

⇒   (a – b)(b – c)(c – a) = ±8

∵   (a – b) + (b – c) + (c – a) = 0

∴   (a – b)³ + (b – c)³ + (c – a)³

       = 3(a – b)(b – c)(c – a) = ±24
Question 8
4 / -1

The coefficient of t⁵⁰ in (1 + t²)²⁵ (1 + t²⁵) (1 + t⁴⁰) (1 + t⁴⁵) (1 + t⁴⁷) is

A
1 + ²⁵C₅

B
1 + ²⁵C₅ + ²⁵C₇

C
1 + ²⁵C₇

D
None of these

Solution

As we are interested in coefficient of t⁵⁰, we shall ignore all the term with exponent more than 50. Thus we can write as

(1 + ²⁵C₁t² + ..... + ²⁵C₂₅t⁵⁰) × (1+t²⁵+t⁴⁰+t⁴⁵+t⁴⁷)

As all the terms in the first have even exponent we can ignore t²⁵, t⁴⁵ and t⁴⁷ too thus coefficient of t⁵⁰ is

= ²⁵C₂₅ + ²⁵C₅ = 1 + ²⁵C₅
Question 9
4 / -1

Directions For Questions

...view full instructions

is a continuous function ∀ x ∈ R and a, b, c are non negative integers then the value of a + b + c is :-

A
5

B
6

C
4

D
7

Solution

ƒ(1^–1) = ƒ(1) = ƒ(1⁺)

⇒ 6a – b – 10c = 3 ......(1)

ƒ(2^–) = ƒ(2) = ƒ(2⁺)

⇒ 5a + 7b + 5c = 37 ......(2)

Solving (1) & (2)

⇒ 16a + 13b = 77 only at a = 4, b = 1, c = 2

⇒ a + b + c = 7
Question 10
4 / -1

ln(f(x)) + g(x) + C

where C is the constant of integration and f (x) is positive, then f (x) + g (x) has the value equal to

A
e^x + sin x + 2x

B
e^x + sin x

C
e^x – sin x

D
e^x + sin x + x

Solution